Problem: $f(x) = \begin{cases} 8 & \text{if } x = 0 \\ 2x^{2}+2 & \text{otherwise} \end{cases}$ What is the range of $f(x)$ ?
Explanation: First consider the behavior for $x \ne 0$ Consider the range of $2x^{2}$ The range of $x^2$ is $\{\, y \mid y \ge 0 \,\}$ Multiplying by $2$ doesn't change the range. To get $2x^{2}+2$ , we add $2$ So the range becomes: $\{\, y \mid y ≥ 2 \,\}$ If $x = 0$ , then $f(x) = 8$ , which eliminates $f(x) = 2$ from the range. The new range is $\{\, y \mid y > 2 \,\}$.